How to delete files if a numerical part of their name is greater than a given number

filesfindrm

I've got files like this:

cap_20151023T122915_791033959.png
cap_20151023T122918_946392456.png
cap_20151023T122920_227637228.png
cap_20151023T122920_875467456.png

and I would like to use the find command to delete those greater than, for example, cap_20151023T122919*, which would result in the deletion of cap_20151023T122920_227637228.png and cap_20151023T122920_875467456.png.

Is there any way to do this, preferably with a single find command?

Best Answer

With zsh and <[x]-[y]> glob operator (matches numbers in the range x to y, inclusive; either of the numbers may be omitted to make the range open-ended) e.g.:

print -rl -- **/cap_20151023T<122920->_*

or, if you want to select only the file names in the 122920-999999 range:

print -rl -- **/cap_20151023T<122920-999999>_*

so with file names like:

tmp/cap_20151023T122915_791033959.png
tmp/cap_20151023T122915791_959.png
tmp/cap_20151023T122918_946392456.png
tmp/cap_20151023T122920_227637228.png
tmp/cap_20151023T1229205_875467456.png
tmp/cap_20151023T122920_875467456.png
tmp/cap_20151023T122980_227637228.png

the first one prints:

tmp/cap_20151023T122915791_959.png
tmp/cap_20151023T122920_227637228.png
tmp/cap_20151023T1229205_875467456.png
tmp/cap_20151023T122920_875467456.png
tmp/cap_20151023T122980_227637228.png

while the second one prints:

tmp/cap_20151023T122920_227637228.png
tmp/cap_20151023T122920_875467456.png
tmp/cap_20151023T122980_227637228.png

If you're happy with the result replace print -rl with rm -f

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i know this is a bit overkill but, this will work every time (even if there are spaces in your filename) and regardless of how file displays the information.

find . -name '*.png' -exec file {} \; | sed 's/\(.*png\): .* \([0-9]* x [0-9]*\).*/\2 \1/' | awk 'int($1) > 500 {print}'

and it prints the dimensions of the picture and the file

explaination:

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Shell – How to delete the files in one folder which are more than 60 days old in UNIX

OK, I have remade this script, and by sorting it backwards it looks like it should work. It compares the year and month to the previous one, and if it is lower it should be the last entry for that month.

#!/bin/bash

#the tac reverses the listing, so we go from newest to oldest, vital for our logic below
FILES=`ls | tac`

#create a cutoff date by taking current date minus our 60 day limit
CUTOFFDATE=`date --date="60 days ago" +%Y%m%d`

#we are setting this value to month 13 of the current year 
LASTYEARMONTH=`date +%Y`13

for file in $FILES; do

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    THISYEARMONTH=${FILEDATE:0:6}

    if [ ! -z $FILEDATE ] && [ $THISYEARMONTH -lt $LASTYEARMONTH ]; then

            echo "$file IS THE LAST FILE OF THE MONTH. NOT DELETING"

    else

            #now we see if the file is older than 60 days (ie, has a LOWER unix timestamp than our cutoff)
            if [ ! -z $FILEDATE ] && [ $FILEDATE -lt $CUTOFFDATE ]; then

                    echo "$file has a file date of $FILEDATE which is older than 60 days."

                    #uncomment this next line to remove
                    #rm $file
            fi
    fi
    LASTYEARMONTH=$THISYEARMONTH
done

Best Answer

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